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I have simple question on C++ linked list. Say I am given a list, I just want to print for each node the value. I found this simple code on the net

while (head!= NULL) {
        cout << head->data;
        head= head->next;
    }

which works just fine.

If I do the following

cout << head->data; 
head= head->next;
cout << head->data << endl;

I get the following error (on LeetCode)

Line 20: Char 25: runtime error: member access within null pointer of type 'ListNode' (solution.cpp) SUMMARY: UndefinedBehaviorSanitizer: undefined-behavior prog_joined.cpp:29:25

I do not understand what I am missing.

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9
  • Actually, you set head=head->next and then use head without checking whether its null or not. If null, using head->data will cause some trouble. Commented Jun 26 at 21:52
  • 1
    My personal choice is fmt::printlin("{}", fmt::join(listname, ", ")); You seem to not want a loop? What would your proposed solution look like if the list contains 3 elements? 200? And given that you're working on a LeetCode solution, are you restricted to the standard library, or is this for personal checks before cleaning up for submission? Commented Jun 26 at 21:53
  • @abcdefg I initially thought that's what was causing the issue but to b emore specific, I do test whether the head is not null and the result is true. So I do not think that is the reason Commented Jun 26 at 21:57
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    It's called dereferencing a null ponter. Your first code sample avoids that. Your second (crashing) code sample fails to avoid that. A null pointer is a pointer that you can't dereference Commented Jun 27 at 2:45
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    Seems someone is trying you to teach you "C"... For crying out loud this is 2025 and std::list has been a part of C++ for ages now. It exists to avoid all the pointer @$## you are running into now Commented Jun 27 at 4:52

2 Answers 2

Reset to default
4 
cout << head->data;          //head could be NULL
head = head->next;           //head->next could be NULL, hence head would be NULL
cout << head->data << endl;  //head could be NULL

As you can see, all three lines can lead to undefined behavior (or a segfault), due to dereferencing a NULL pointer.

Whereas the previous snippet:

while (head != NULL) {
    cout << head->data;
    head = head->next;
}

is safe, because you always have a test if head == NULL, and therefore never reach the point where the members of head are accessed (head->data, head->next), needless to say that the entire block is skipped if the test fails (if head == NULL).

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3 Comments

So, the key takeaway is always validating pointer before dereference. And using the while(head) loop is the simplest and most idiomatic pattern for traversing and printing linked lists?
@Hawk whenever you dereference a pointer, the pointer MUST be non-null. So yes, it should be tested all the time. But in a synchronize function call, you may just test it in the beginnin.
@Hawk Yes, validating pointers (before dereferencing) is a must do. When a pointer is passed as argument to a function, one would first test (assert) for NULL. In case of list items, you'll probably find something like this: for (item_ptr item = head; item; item = item->next) {...}, but in modern C++, one would typically use iterators and test if some iterator has reached the end marker (e.g. pos != end), and there are still all algorithm functions, which essentially makes manual iteration obsolete.
1

You are not checking if head is NULL before accessing its data.

In the first code snippet there is a check that validates that head is not NULL before accessing its data.

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2 Comments

The same comment was raised by someone else and I do not think this is the issue.
The error message said exactly that "member access within null pointer of type 'ListNode' (solution.cpp) SUMMARY: UndefinedBehaviorSanitizer: undefined-behavior prog_joined.cpp:29:25" Which was exactly that. Your code attempted to dereference a null pointer on line 29 of solution.cpp.

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